TITLE: Find a closed form of the power series
QUESTION [9 upvotes]: Let a power series $$S(x)=\sum_{n=1}^{\infty}\frac{x^{n}}{4n+1},$$ then  $1$ is the radius of convergence of $S$ .In fact $S(x)$ convergens for each $x\in[-1,1).$
My work is to find a closed form of this power series.

The following is my solution:
$$S(x)=\sum_{n=1}^{\infty}\frac{(4n+1)x^{n}-4nx^{n}}{4n+1}=\sum_{n=1}^{\infty}x^{n}-4x\sum_{n=1}^{\infty}(\frac{x^{n}}{4n+1})^{'}.$$then we have $$S(x)+4x S^{'}(x)=\frac{x}{1-x}.(\text{a first-order linear differential equation })\Rightarrow $$ 
$$\mathbf{A.}\quad 
S(x)=\frac{1}{x^{\frac{1}{4}}}[C_{1}+\frac{\ln(1+x^{\frac{1}{4}})+2\arctan(x^{\frac{1}{4}})-\ln(1-x^{\frac{1}{4}})-4x^{\frac{1}{4}}}{4}](1>x>0);$$$$\mathbf{B.}\quad
 S(x)=-\frac{1}{8(-x)^{\frac{1}{4}}}\begin{Bmatrix}
C_{2}+8(-x)^{\frac{1}{4}}\\+2\sqrt{2}\arctan[1-\sqrt{2}(-x)^{\frac{1}{4}}]-2\sqrt{2}\arctan[1+\sqrt{2}(-x)^{\frac{1}{4}}]\\+
\sqrt{2}\ln[1-\sqrt{2}(-x)^{\frac{1}{4}}+\sqrt{-x}]-\sqrt{2}\ln[1+\sqrt{2}(-x)^{\frac{1}{4}}+\sqrt{-x}]
\end{Bmatrix}\quad(-1\leq x<0).$$ 
Since $S(0)=0$ and $S(x)$ is continuous at $x=0$ , I conclude that $C_{1}=C_{2}=0.$

I am not sure my solution is right .I need someone to check my answer,or If you   have some good ideas about how to solve this question ,please give me some hints.Any help is going to be appreciated!

REPLY [15 votes]: Hint. You may consider the following derivative 
$$
\left(\sum_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1}\right)'=\sum_{n=0}^{\infty}x^{4n}=\frac1{1-x^4},\quad |x|<1.
$$ Then, integrating the power series termwise  for $|x|<1$ (which is allowed), you get
$$
\sum_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1}=\int_0^x\frac1{1-t^4}dt=\int_0^x\left(\frac{1}{4 (1-t)}+\frac{1}{4 (1+t)}+\frac{1}{2 \left(1+t^2\right)}\right)dt
$$ giving easily

$$
\sum_{n=\color{red}{1}}^{\infty}\frac{x^{4n+1}}{4n+1}=\frac14 \ln \left(\frac{1+x}{1-x} \right)+\frac12\arctan x-x, \quad |x|<1,
$$ 

from wich you deduce your initial series by dividing first by $x$ then putting $x \to x^{1/4}$.